Energy Analysis of Closed Systems

Chapter 4

Recall that a closed system does not include mass transfer

Heat can get in or out Work can get in or out Matter does not cross

the system boundaries

Mechanical WorkThere are many kinds of mechanical workThe most important for us will be moving

boundary work Wb

Sometimes called PdV work

The primary form of work involved in automobile engines is moving boundary work produced by piston cylinder devices

The mass of the substance contained within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and make it move.

Boundary Work

W W FdsF

AAds

PdV

b b

z z zz

1

2

1

2

1

2

1

2

The area under the curve is the work

The work depends on the process path

Consider some special cases

Constant volumeConstant pressureIsothermal for an ideal gasPolytropic

Constant Volume

P

V

1

2

W PdVb z12 0

Constant Pressure

P

V

1 2

W

W PdV P dV P V Vb z z1

2

1

2

2 1b g

Isothermal for an ideal gas

P

V

1

2

PmRT

V

W PdVmRT

VdV

mRTV

V

b

FHG

IKJ

z z1

2

1

2

2

1

ln

Polytropic

P

V

1

2

PV n constant

W PdVConst

VdV

PV PV

n

PVV

V

b n

FHG

IKJ

z z1

2

1

2

2 2 1 1

2

1

1,

ln ,

n 1

= n = 1

n=1

PV1= constant equivalent to the isothermal case for an ideal gas

PV= mRT

2

1

lnb

VW mRT

V

2

1

lnV

PVV

P

V

Lets go through the polytropic case integration step-wise

CPV n

nV

CP nCV

2

1

2

1dVCVPdVW n

b

1

1

n

VCW

n

b

2

1

1

11

12

n

VVC

nn

But…

nV

CP

22 And…. nV

CP

11

1

11

12

n

VVC

nn

Wb

So…

1122 VPVPWb 1 n

nn V

CV

V

CV

1

1

2

2

1 n

So far, we have not assumed an ideal gas in this derivation, if we do, then….

mRTPV

n

mRTmRTWb

112

n

TTmRWb

112

Energy Balance for a Closed System (Chapter 2)

E E Ein out system How can energy get in and out of a closed system?

Heat and Work

Total Energy entering the system

Total Energy leaving the system

The change in total energy of the system

- =

Rate form

/in out systemE E E dE dt

Q W Enet net system

E = U + KE + PE

E U KE PE

0 0If the system isn’t moving

Q W U KE PEnet net

In a cyclic process, one where you end up back where you started:

0E0 WQ

net in net outQ W

You convert heat to work, or vise versa

The net work is the area inside the figure

1

2

Pre

ssure

Volume

Wnet

If we can calculate the work done for each step in this process, we can find the net work produced or consumed by the system

Calculating Properties

Chapter 4b

We know it takes more energy to warm up some materials than others

For example, it takes about ten times as much energy to warm up a pound of water, as it does to warm up the same mass of iron.

Specific Heats – Cp and Cv

Also called the heat capacityEnergy required to raise the

temperature of a unit mass one degree

UnitskJ/(kg 0C) or kJ/(kg K) cal/(g 0C) or cal/(g K) Btu/(lbm 0F) or Btu/(lbm 0R)

Consider a stationary constant volume system

E=U+KE +PE

ddU

UU

mCvdT

duCvdT

vv T

uC

Q-W=ΔU

Q

First Law

Consider a stationary constant pressure system

It takes more energy to warm up a constant pressure system, because the system boundaries expand

You need to provide the energy to increase the internal energy do the work required to move the system

boundary

Consider a stationary constant pressure system

E=U+KE +PE

Q=ΔU+PΔVΔH

UU

mCpdTdhCpdT

pp

hC

T

Q-W=ΔU

Q

Q-PΔV=ΔU

dH

Cp is always bigger than Cv

pp T

hC

h includes the internal

energy and the work required to expand the system boundaries

Cp and Cv are properties

Both are expressed in terms of u or h, and T, which are properties

Because they are properties, they are independent of the process!!

The constant volume or constant pressure process defines how they are measured, but they can be used in lots of applications

Ideal Gases

RTPv

Tuu

Joule determined that internal energy for an ideal gas is only a function of temperature

Pvuh

RTPv

RTuh Which means that h is also only a function of temperature for ideal gases!!

For non ideal gases both h and u vary with the state

Specific Heats vary with temperature – but only with temperature –for an ideal gas

Note that the Noble gases have constant specific heats

Why is water on this chart?

dTTCdu v )(

dTTCuuu v2

112

dTTCdh p )(

dTTChhh p2

112

If Cv and Cp are functions of temperature – how can we integrate to find U and H?

Use an average value, and let the heat capacity be a constant

2

1 12,12 TTCdTTCuuu avevv

2

1 12,12 TTCdTTChhh avepp

That only works, if the value of heat capacity changes linearly in the range you are interested in.

OK Approximation

Crummy Approximation

Sometimes the best you can do is the room temperature value

What if you need a better approximation?

All of these functions have been modeled using the form

Cp = a + bT + cT2 + dT3

The values of the constants are in the appendix of our book – Table A-2c

2

1

322

1)( dTdTcTbTadTCh p

2 2 3 3 4 42 1 2 1 2 1

2 1( )2 3 4

b T T c T T d T Th a T T

This is a pain in the neck!! Only do it if you really need to be very accurate!!

Isn’t there a better way?

Use the Ideal Gas Tables

Table A-17 pg 910 (air)Both u and h are only functions of T –

not pressureRelative values have been tabulated

for many ideal gasesIf the gas isn’t ideal – then it’s a

function of both T and P and these tables don’t work!!

Cp is modeled in the Appendix as a function of temperature – so you could calculate h, but what if you want to calculate u? You’d need Cv

There is no corresponding Cv table !!

Cp = Cv + R

Three Ways to Calculate u

v

p

C

Ck

Specific Heat Ratio

k does not vary as strongly with temperature as the heat capacity

k = 1.4 for diatomic gases (like air)

k = 1.667 for noble gases

Use in Chapter 7

Solids and Liquids

Treat as incompressible fluids

Cp = Cv = C

du C dT CdTV

u C T C T T ( )2 1

h u Pv

dh du Pdv vdP

h u v P C T v P

But dv is 0 if the system is incompressible

0

small

Summary Boundary work

Looked at four different special cases Constant volume Constant pressure Constant temperature Polytropic

2

1bW PdV

Summary

First Law for a closed system

Q W U KE PEnet net

Summary

Defined Constant pressure heat capacity Constant temperature heat capacity

By performing a first law analysis of a closed system

pp T

hC

vv

uC

T

SummaryThe three ways to calculate changes in

Internal energy (u)Enthalpy (h)

Look up properties at state one and two in the tables

Assume constant values of specific heat, then integrate

Use the curve fit equation for specific heat, then integrate